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X = 2F + 3Fcos™ - 2F cosf -F -66
tan 0 = - V3.
Therefore the resultant force has a magnitude 2 F and makes an angle of —60° with the :i*-axis.
Taking the moments about an axis through the center of the hexagon, we obtain
FIG. 36.
11 Fa,
therefore          D = 5.5 a,
where a is the distance of the center from the lines of action of the forces.
56.  Resultant of a System of Parallel Forces. — Let R be the
resultant of the parallel forces Fi, F2, . . . , Fn; which act upon a rigid body. Then, since the forces are parallel, the resultant force equals the algebraic sum of the given forces. Thus
R=Fl + F,+ • - - +Fn,
and                 RD = Fidi + FA+ - • • +Fndn.
Now take the s-axis parallel to the forces and let x{ and y{ denote the distances of F* from the yz-plane and the xz-plane, respectively. Then the last equation may be split into two parts, one of which gives the moments about the and the other about the y-axis. Thus,
Ry = FM + F*y* + - - - + Fnyn, }
where x and y are the coordinates of the point in the xy-plane through which the resultant force passes.    In other